How do you use the Ratio Test on the series #sum_(n=1)^oo(-10)^n/(4^(2n+1)(n+1))# ?

Question

9997 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2014-09-04T00:24:27

By Ratio Test, the posted series converges absolutely.

By Ratio Test:

#lim_{n to infty}|a_{n+1}/a_n|=lim_{n to infty}|{(-10)^{n+1}}/{4^{2n+3}(n+2)}cdot{4^{2n+1}(n+1)}/{(-10)^n}|#

By canceling out common factors:

#=lim_{n to infty}|{-10(n+1)}/{4^2(n+2)}|#

since #|{-10}/4^2|=5/8#, we have:

#=5/8lim_{n to infty}(n+1)/(n+2)#

by dividing the numerator and the denominator by #n#,

#=5/8 lim_{n to infty}{1+1/n}/{1+2/n}=5/8cdot 1=5/8<1#

Hence, #sum_{n=1}^{infty}{(-10)^n}/{4^{2n+1}(n+1)}# is absolutely convergent.